将空字符串传递给正则表达式对象时,搜索结果是匹配对象而非None.应该是没有,因为没有什么可以匹配的？
import re
m = re.search("", "some text")
if m is None:
print "Returned None"
else:
print "Return a match"
顺便说一句,使用特殊符号^并$产生相同的结果.
由于在我的工件设置中丢失了管理员用户凭据，我被卡住了。
尝试过的步骤在
https://www.mail-archive.com/artifactory-users@lists.sourceforge.net/msg01929.html
http://forums.jfrog.org/Lost-admin-password-td5435810.html
但仍然没有成功。日志说：
[http-bio-8081-exec-3] [ERROR] (o.a.w.DefaultExceptionMapper:105) - Connection lost, give up responding.
org.apache.wicket.protocol.http.servlet.ResponseIOException: ClientAbortException: java.net.SocketException: Connection reset by peer: socket write error
at org.apache.wicket.protocol.http.servlet.ServletWebResponse.flush(ServletWebResponse.java:255) ~[wicket-core-1.5.3.jar:1.5.3]
at org.artifactory.webapp.wicket.application.HeaderBufferingWebResponse.flush(HeaderBufferingWebResponse.java:89) ~[artifactory-web-application-3.0.2.jar:na]
at org.artifactory.webapp.wicket.application.IgnoreEofWebResponse.flush(IgnoreEofWebResponse.java:105) ~[artifactory-web-application-3.0.2.jar:na]
at org.apache.wicket.request.resource.AbstractResource.setResponseHeaders(AbstractResource.java:611) ~[wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.resource.AbstractResource.respond(AbstractResource.java:485) ~[wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.handler.resource.ResourceRequestHandler.respond(ResourceRequestHandler.java:77) ~[wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.handler.resource.ResourceReferenceRequestHandler.respond(ResourceReferenceRequestHandler.java:105) ~[wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.cycle.RequestCycle$HandlerExecutor.respond(RequestCycle.java:750) ~[wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.RequestHandlerStack.execute(RequestHandlerStack.java:64) ~[wicket-request-1.5.3.jar:1.5.3]
at org.apache.wicket.request.cycle.RequestCycle.execute(RequestCycle.java:252) [wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.cycle.RequestCycle.processRequest(RequestCycle.java:209) [wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.request.cycle.RequestCycle.processRequestAndDetach(RequestCycle.java:280) [wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.protocol.http.WicketFilter.processRequest(WicketFilter.java:162) [wicket-core-1.5.3.jar:1.5.3]
at org.apache.wicket.protocol.http.WicketFilter.doFilter(WicketFilter.java:218) [wicket-core-1.5.3.jar:1.5.3]
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:243) [catalina.jar:7.0.39]
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210) [catalina.jar:7.0.39]
at org.artifactory.webapp.servlet.RepoFilter.execute(RepoFilter.java:166) [artifactory-web-application-3.0.2.jar:na]
at org.artifactory.webapp.servlet.RepoFilter.doFilter(RepoFilter.java:85) [artifactory-web-application-3.0.2.jar:na]
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:243) …多达255个,我能理解整数如何被存储在char和unsigned char;
#include<stdio.h>
int main()
{
unsigned char a = 256;
printf("%d\n",a);
return(0);
}
在上面的代码中,对于unsigned char和char,我的输出为0.
对于256我认为这是整数存储在代码中的方式(这只是猜测):
第一个256转换为二进制表示,即100000000(总共9位).
然后他们删除删除最左边的位(设置的位),因为char数据类型只有8位内存.
因此它在内存中存储为00000000,这就是为什么它的打印0作为输出.
猜测是否正确或有任何其他解释？
正在与向我发送没有内容长度标头但有内容的 HTTP 请求的客户端进行测试。
如何在没有 contentlength 标头的帮助下提取此内容？
假设我有一个void*内存地址,我需要打印位于该内存地址中的位.我怎样才能做到这一点？
在我的处理器中,内存地址是32位,内存值也是32位,int也是32位.所以我想这样做:
unsigned int value = *memory_address;
然后通过简单的算术(一些mod和div操作)来获取保存的值的位memory_address.
例如,value mod 2将给出该值的最后一位,依此类推.但是从我所知道的(我期待不同的位)它不起作用.有什么想法吗？
还有,是否有人知道现成的C源代码"做"这样,从内存中读取/写入位？
我在其中有以下类和方法
public class A extends B implements C{
public void validateTicketGrantingTicket(final TicketGrantingTicket ticketGrantingTicket) throws InvalidTicketException {
if (ticketGrantingTicket != null)
{
if (!ticketGrantingTicket.getHostDomain().equalsIgnoreCase(getServerName()))
{
throw new InvalidTicketException();
}
}
}
public String getServerName()
{
String serverName = "";
HttpServletRequest request = ((ServletRequestAttributes) RequestContextHolder.getRequestAttributes()).getRequest();
if (request != null)
{
serverName = request.getServerName().toLowerCase();
}
return serverName;
}
}
现在我正在写ATest课并嘲笑A班.
public class ATest {
private A a;
@Before
public void init(){
A = mock(A.class);
when(A.getServerName()).thenReturn("phoenix.edu.abc");
}
@Test
public void validateTicketGrantingTicketTest() throws InvalidTicketException{
a …我有个问题。我想在 kubernetes 上安装 helm 但是当我想运行这个命令helm init --upgrade但我有这个错误：
Creating /root/.helm
Creating /root/.helm/repository
Creating /root/.helm/repository/cache
Creating /root/.helm/repository/local
Creating /root/.helm/plugins
Creating /root/.helm/starters
Creating /root/.helm/repository/repositories.yaml
Error: Looks like "https://kubernetes-charts.storage.googleapis.com" is not
a valid chart repository or cannot be reached: Get https://kubernetes-
charts.storage.googleapis.com/index.yaml: dial tcp 216.58.197.80:443: i/o
timeout
我想代理设置没有设置，但我不知道如何去做。
一个主意 ？
感谢您的帮助，
真挚地，
Killer_Minet
我在 minikube 上运行 kubernetes，我在代理后面，所以我在 /etc/systemd/system/docker.service.d/http-proxy.conf 中为 docker 设置了 env 变量（HTTP_PROXY & NO_PROXY）。我能够做 docker pull 但是当我运行下面的例子时
kubectl run hello-minikube --image=gcr.io/google_containers/echoserver:1.4 --port=8080
kubectl expose deployment hello-minikube --type=NodePort
kubectl get pod
pod 永远不会启动，我收到错误消息
desc = unable to pull sandbox image \"gcr.io/google_containers/pause-amd64:3.0\"
docker pull gcr.io/google_containers/echoserver:1.4 工作正常
我会直截了当:我有一个这样的字符串(但有数千行)
Ach-emos_2
Ach. emos_54
Ach?mos_18
?žuolas_4
Somtehing else_2
我需要删除不符合行a-z和?????š??ž加上_加any integer(第三和第四线匹配这一点).这应该是不区分大小写的.我认为正则表达式应该是
[a-z?????š??ž]+_\d+ #don't know where to put case insensitive modifier
但是,应该如何看待匹配非alpha(和立陶宛字母)加上下划线加整数的行的正则表达式？我试过了
re.sub(r'[^a-z?????š??ž]+_\d+\n', '', words)
但没有好处.
提前谢谢,对不起,如果我的英语不太好.
我正在尝试学习C中的指针并且正在编写这个小整数数组指针练习,但遇到了一个无效的sizeof类型int[]问题的应用程序.请告诉我哪里出错了以及如何解决.谢谢.
#include <stdio.h>
int intA[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int intB[];
void int_copy(const int *source, int *destionation, int nbr)
{
int i;
for(i=0;i<nbr;i++)
{
*destionation++ = *source++;
}
}
int main()
{
int *ptrA = intA;
int *ptrB = intB;
int sizeA = sizeof(intA);
int nbrA = sizeof(intA)/sizeof(int);
printf("\n\n");
printf("[Debug]The size of intA is:%d\n", sizeA);
printf("[Debug]That means the number of elements is:%d\n", nbrA);
printf("\n\nThe values of intA are:\n");
int i; …当我尝试从其他函数调用函数时(我确实在if语句中),我一直收到错误.喜欢:
def function1(num):
num = 5
if num == 5:
function2("This is 5")
return
def function2(x):
print x
return
我以前没能找到像这样的问题.它可能吗？
问题是,当我的句子中包含两个字母数量相同的单词时,如何让程序在阅读时给出第一个最长的单词而不是两个单词？
import sys
inputsentence = input("Enter a sentence and I will find the longest word: ").split()
longestwords = []
for word in inputsentence:
if len(word) == len(max(inputsentence, key=len)):
longestwords.append(word)
print ("the longest word in the sentence is:",longestwords)
例如:快速的棕色狐狸......现在程序给了我"快速"和"棕色",如何调整我的代码只是给我"快速",因为它是第一个最长的单词？
好吧,我是python的新手,我正在学校上课,我对此有点困惑.我们正在编写一个程序/脚本来计算买卖股票的交易,由于某种原因,我无法获得"余额"变量随着时间的推移积累并从买卖股票中减去并增加.这是代码.任何输入都会很棒^.^
def main():
#Below is the variables used in the context
balance = float(input("Starting cash amount? "))
numtrades = int(input("Number of trades for today?"))
print('You do not own any shares, but have', balance, 'in cash')
buy = 0
sell = 0
price_per_share = 0
transaction_amount = 0
transaction_amount_buy = 0
transaction_amount_sell = 0
#Below is the prompt for the first transaction, which requires a buy
num_shares = int(input('Number of shares to buy?'))
price_per_share = float(input('Price per share?'))
print(num_shares, "shares for …